Problem 34 - University Physics with Modern Physics

Question

An x-ray tube is operating at voltage \(V\) and current \(I\). (a) If only a fraction \(p\) of the electric power supplied is converted into x rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in J/kg \(\bullet\) K), at what average rate would its temperature rise if there were no thermal losses? (c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at 18.0 kV and 60.0 mA that converts 1.0\(\%\) of the electric power into x rays. Assume that the 0.250-kg target is made of lead (\(c\) = 130 J/kg \(\bullet\) K). (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

Step-by-Step Solution

Answer

Energy as x-rays is 10.8 W; target's temperature rises at 204 °C/hour; suitable targets are tungsten or lead.

Step 1: Understanding Electric Power Supplied

The electric power supplied to the x-ray tube is given by the product of the voltage \( V \) and current \( I \). Therefore, the power supplied \( P \) is \( P = V \times I \).

Step 2: Calculating Energy Delivered to Target

Out of the total power supplied, only a fraction \( p \) is converted into x-rays. Thus, the energy (power) being delivered to the target as x-ray energy is \( P_{xray} = p \times P = p \times V \times I \).

Step 3: Calculating Rate of Temperature Rise

The remaining power \( (1 - p) \times P \) causes the target's temperature to rise. If we assume no thermal losses, the rate of temperature increase \( \Delta T/\Delta t \) can be calculated using the formula: \( (1 - p) \times V \times I = m \times c \times \frac{\Delta T}{\Delta t} \).

Step 4: Calculating Numerical Values

First, convert all units to consistent SI units: 18.0 kV is 18000 V, and 60.0 mA is 0.060 A. The power supplied is \( P = 18000 \times 0.060 \) W. The power converted to x-rays is \( 0.01 \times P \), and the power causing temperature rise is \( 0.99 \times P \). For the temperature, \( \frac{\Delta T}{\Delta t} = \frac{0.99 \times 18000 \times 0.060}{0.250 \times 130} \).

Step 5: Evaluating Physical Properties of Target Material

For a practical target, material must withstand high temperatures, have a suitable thermal conductivity, and not deteriorate quickly. Ideal elements include tungsten and lead due to high melting points and thermal stability.

Key Concepts

Electric Power

In an x-ray tube, electric power is the key element that initiates the production of x-rays. This power is defined as the product of voltage (V) and current (I), giving us the formula: \( P = V \times I \). Understanding this is crucial because it represents the total energy input into the x-ray tube.
When you apply a voltage across a current, the energy is harnessed to generate x-rays. This energy transfer, however, is not perfect. Only a small fraction of the supplied electric power is utilized to create x-rays, with a significant portion generating unwanted heat.
The understanding of electric power helps in calculating how much energy is actually converted into useful x-ray output versus how much contributes to heat and potential overheating issues within the tube, which leads us into the next concept.

Energy Conversion

Energy conversion in an x-ray tube involves transforming electric energy into x-rays and heat. The efficiency of this conversion process is essential for the performance and operational safety of x-ray equipment.
Here’s how it works:

• A given ratio \( p \) (usually a small percentage) of the electric power is converted into x-ray photons.
• The power converted into x-rays is calculated using \( P_{xray} = p \times V \times I \).
• The remaining power, \( (1 - p) \times V \times I \), is largely dissipated as heat.

This inefficient conversion highlights the need for developing materials and technologies that can better utilize the supplied electric power for x-ray generation.
Understanding these principles aids in recognizing the thermal challenges faced by the target within the x-ray tube, leading us to the next concept.

Temperature Rise

The leftover energy from the electric power in an x-ray tube, not converted into x-rays, is transformed into heat. This results in the temperature rise of the target material. Understanding this phenomenon is critical in ensuring the durability and reliability of x-ray tubes.
The temperature rise rate \( \frac{\Delta T}{\Delta t} \) can be calculated by equating the heat energy to the product of mass (m), specific heat (c), and the temperature change over time:\[ (1 - p) \times V \times I = m \times c \times \frac{\Delta T}{\Delta t} \]Given that even a small efficiency loss can lead to significant heat generation, this calculation provides insight into how quickly the target will heat up.
Managing temperature rise is essential to prevent damage and maintain the x-ray tube's effectiveness, which ties into the significance of target material properties.

Target Material Properties

Selecting the proper target material in an x-ray tube is pivotal to manage high temperatures resulting from energy conversion inefficiency. Not all materials can endure the extreme conditions within an x-ray tube environment, which makes choosing robust materials necessary.
Key considerations for target materials include:

• High melting point to withstand intense heat without melting.
• Good thermal conductivity to dissipate heat efficiently and prevent overheating.
• Material stability under constant bombardment of electrons and the resulting heating cycle.

Metals like tungsten and lead are commonly used due to their ability to meet these criteria successfully. Tungsten's high melting point and durability make it a popular choice in high-performance x-ray tubes.
These considerations ensure that the target material can effectively manage temperature rise, helping in maintaining the x-ray tube's performance and longevity.